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Homework Help: Minimizing Volume

  1. Oct 20, 2004 #1
    I'm supposed to find the linear plane which cuts through first octant and results in the minimum volume underneath the plane. The plane must pass under (2,3,4). I think I need to use Lagrangian multipliers and minimize V=(1/2)xyz.. I'm not sure what the constraint would be though
  2. jcsd
  3. Oct 24, 2004 #2
    ok, after thinking about this some more...

    x/a+y/b+z/c=1 will give me a plane where the x, y, z intercepts will be a, b, and c respectively. Therefore if the point passes through 2, 3, 4, then the constraint will have to be 2/a+3/b+4/c = 1

    and then V will just be (1/2)abc

    and then just apply the lagrangian method

    Does that sound like a good way to approach this problem?
  4. Oct 25, 2004 #3
    sounds good to me
  5. Oct 25, 2004 #4


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    I don't see how this can HAVE a solution. If your purpose is to get minimum volume and your only requirement is that the plane must "pass under (2,3,4), can't you make the volume arbitrarily small by making the intercepts, x, y, z, arbitrarily close to 0?
  6. Oct 25, 2004 #5
    I'm sure he meant that it goes THROUGH the point :)
  7. Oct 25, 2004 #6
    yeah thats what I meant lol my mistake
  8. Oct 26, 2004 #7


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    Then the way you suggested orginally is the way to go. The volume you want to maximize would be abc, of course, and the constraint is 2/a+ 3/b+ 4/c= 1.
    Taking the grad of abc give <bc, ac, ab> while taking the grad of (2/a+ 3/b+ 4/c) gives <-2/a2[/sub], -3/b2, -4/c2>.

    At the values of a,b,c that maximize the volume while satisfying the constraint, we must have <bc, ac, ab>= &lambda;<-2/a2[/sub], -3/b2, -4/c2> for some &lambda;

    That is, bc= -2&lambda;/a2[/sub], ac= -3&lambda;/b2[/sub], ab= -4&lambda;/c2[/sub].

    Divide the first equation by the second to get
    [tex]\frac{bc}{ac}= \frac{-2\lambda}{a^2}\frac{b^2}{3\lambda}[/tex]
    [tex]\frac{b}{a}= \frac{2b^2}{3a^2}[/tex]
    which is the same as 3a= 2b.

    Similarly, dividing the second equation by the third gives
    [tex]\frac{c}{b}= \frac{3c^2}{4b^2}[/tex]
    which is the same as 4b= 3c.
    That is, b= (3/2)a and c= (4/3)b= 2a.
    Now put that into 2/a+ 3/b+ 4/c= 1 to solve for a, b, c.
  9. Oct 26, 2004 #8
    This is what I did, I got -2λ/(a^2) = bc, -3λ/(b^2) = ac, -4λ/(c^2)=ab. Mutiply each by a, b, and c respectively and so: -2/a = -3/b = -4/c (take out negative and plug into constraint)
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